Why do we need spontaneous symmetry breaking in Lagrangian formalism?

I have always struggled with the concept of spontaneous symmetry breaking. It seems to me that many others don't find it very intuitive as well, but that could be just me having difficulties with the big picture behind it.

Anyway, I would like to change it and understand this concept better, if it's possible. To be honest, without any formulas and Lagrangians, to me it just looks like this:

1. Write the 'wrong' Lagrangian in which particles do not have mass,
2. From experiment find out that particles actually do have mass,
3. Fix the initial Lagrangian by inventing new particles and connecting them to the old particles in a way that they now have mass (part where we 'generate' mass),
4. Call the Lagrangian 1. and 3. related by symmetry breaking/symmetry restauration.

So, why do 1-4? Couldn't one immediately write the 'effective' Lagrangian with all the desired masses and work with that? Why complicate if it can be simpler?

• You can do write an effective lagrangian with mass terms, but it turns out that they destroy gauge invariance. SSB turned out to be a good mechanism to dynamically produce masses in a gauge-covariant way. – Phoenix87 Dec 27 '14 at 20:01

We don't need "spontaneous symmetry breaking in Lagrangian formalism" in general.
We can use spontaneous symmetry breaking to generate masses for some fields.

It so turns out that spontaneous symmetry breaking introduces those masses without
breaking some other nice features of the model at hand: mostly renormalizability and
that symmetry that we are breaking.

It also so turns out that that is exactly what we need to describe objective reality.

• So, for example, in the famous case of the chiral symmetry breaking, we have the following situation: Chiral symmetry breaking can be explained in terms of the sigma model Lagrangian: $$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi)^2 - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4$$, and this is the effective description of some system, involving a scalar field $\phi$. This Lagrangian describes an ordinary scalar field theory. Now, let's say that in some cases, for some reason, the mass term of this field becomes negative, $m^2 < 0$. – Dee Dec 28 '14 at 15:49
• In the chiral symmetry case, this for some critical value of temperature, because $m = m(T)$ and it changes value for $T < T_C$ This is the moment where the spontaneous symmetry breaking occurs. Now something funny happens with the field $\phi$, because we cannot treat it as a small excitation around the $\phi = 0$ vaccum, since this is not the minimum of the potential anymore (if the mass term is negative). This is the moment where our perturbative approach breaks. In order to have a perturbative quantum field theory we have to consider exctiations around true vacuum. – Dee Dec 28 '14 at 15:51
• So, now we have to invent a new field: $$\phi \rightarrow \langle \phi \rangle + \phi'$$, and we call $\phi'$ a physical field. The inital Lagrangian can now be re-written as: $$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi')^2 + \frac{3m^4}{2\lambda} - m^2\phi'^2 - \sqrt{\frac{\lambda}{6}}m\phi'^3 - \frac{\lambda}{4!}\phi'^4$$. Since the field $\phi'$ has positive mass term, this theory is now good for perturbative approach. In the case of the chiral symmetry story, the field $\phi'$ are physical pions and sigma meson $$\phi' = (\vec{\pi}, \sigma)$$ i.e. the real particles from nature. – Dee Dec 28 '14 at 15:53
• My question is, was the old Lagrangian good theory or not? If it couldn't describe experiment very well, then why keep it, and patch it to the real Lagrangian with some complicated mechanism? – Dee Dec 28 '14 at 15:54
• No, @Dee, stackexchange doesn't work like that. It's not a discussion forum. There's no chit-chat. Please, take the tour: stackoverflow.com/tour And next time put more effort in the question -- not in the comments on an answer. – Kostya Dec 28 '14 at 20:44